Integrand size = 21, antiderivative size = 115 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=2 a b \left (a^2+2 b^2\right ) x+\frac {b^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d} \]
2*a*b*(a^2+2*b^2)*x+b^4*arctanh(sin(d*x+c))/d+1/3*a^2*(2*a^2+17*b^2)*sin(d *x+c)/d+4/3*a^3*b*cos(d*x+c)*sin(d*x+c)/d+1/3*a^2*cos(d*x+c)^2*(a+b*sec(d* x+c))^2*sin(d*x+c)/d
Time = 0.40 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {24 a b \left (a^2+2 b^2\right ) (c+d x)-12 b^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 a^2 \left (a^2+8 b^2\right ) \sin (c+d x)+12 a^3 b \sin (2 (c+d x))+a^4 \sin (3 (c+d x))}{12 d} \]
(24*a*b*(a^2 + 2*b^2)*(c + d*x) - 12*b^4*Log[Cos[(c + d*x)/2] - Sin[(c + d *x)/2]] + 12*b^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*a^2*(a^2 + 8 *b^2)*Sin[c + d*x] + 12*a^3*b*Sin[2*(c + d*x)] + a^4*Sin[3*(c + d*x)])/(12 *d)
Time = 0.84 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4328, 3042, 4562, 27, 3042, 4535, 24, 3042, 4533, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 \sec ^2(c+d x) b^3+8 a^2 b+a \left (2 a^2+9 b^2\right ) \sec (c+d x)\right )dx+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+8 a^2 b+a \left (2 a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{3} \left (\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}-\frac {1}{2} \int -2 \cos (c+d x) \left (3 \sec ^2(c+d x) b^4+6 a \left (a^2+2 b^2\right ) \sec (c+d x) b+a^2 \left (2 a^2+17 b^2\right )\right )dx\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) \left (3 \sec ^2(c+d x) b^4+6 a \left (a^2+2 b^2\right ) \sec (c+d x) b+a^2 \left (2 a^2+17 b^2\right )\right )dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+6 a \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b+a^2 \left (2 a^2+17 b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{3} \left (6 a b \left (a^2+2 b^2\right ) \int 1dx+\int \cos (c+d x) \left (3 \sec ^2(c+d x) b^4+a^2 \left (2 a^2+17 b^2\right )\right )dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) \left (3 \sec ^2(c+d x) b^4+a^2 \left (2 a^2+17 b^2\right )\right )dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}+6 a b x \left (a^2+2 b^2\right )\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+a^2 \left (2 a^2+17 b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}+6 a b x \left (a^2+2 b^2\right )\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{3} \left (3 b^4 \int \sec (c+d x)dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{d}+6 a b x \left (a^2+2 b^2\right )\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 b^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{d}+6 a b x \left (a^2+2 b^2\right )\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{d}+6 a b x \left (a^2+2 b^2\right )+\frac {3 b^4 \text {arctanh}(\sin (c+d x))}{d}\right )\) |
(a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + (6*a*b*(a ^2 + 2*b^2)*x + (3*b^4*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Si n[c + d*x])/d + (4*a^3*b*Cos[c + d*x]*Sin[c + d*x])/d)/3
3.5.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* (n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Time = 0.76 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.85
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{2} b^{2} \sin \left (d x +c \right )+4 a \,b^{3} \left (d x +c \right )+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(98\) |
| default | \(\frac {\frac {a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{2} b^{2} \sin \left (d x +c \right )+4 a \,b^{3} \left (d x +c \right )+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(98\) |
| parallelrisch | \(\frac {24 a^{3} b x d +48 a \,b^{3} d x +12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+a^{4} \sin \left (3 d x +3 c \right )+12 a^{3} b \sin \left (2 d x +2 c \right )+9 a^{4} \sin \left (d x +c \right )+72 a^{2} b^{2} \sin \left (d x +c \right )}{12 d}\) | \(110\) |
| risch | \(2 a^{3} b x +4 a \,b^{3} x -\frac {3 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b^{2}}{d}+\frac {3 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{4}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{4}}{d}+\frac {a^{4} \sin \left (3 d x +3 c \right )}{12 d}+\frac {a^{3} b \sin \left (2 d x +2 c \right )}{d}\) | \(169\) |
| norman | \(\frac {\left (-2 a^{3} b -4 a \,b^{3}\right ) x +\left (-6 a^{3} b -12 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (2 a^{3} b +4 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (6 a^{3} b +12 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {4 a^{2} \left (a^{2}-2 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{2} \left (a^{2}-2 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {4 a^{2} \left (a^{2}+2 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 a^{2} \left (a^{2}+2 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (7 a^{2}-18 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {2 a^{2} \left (7 a^{2}+18 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(360\) |
1/d*(1/3*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+4*a^3*b*(1/2*cos(d*x+c)*sin(d*x+c )+1/2*d*x+1/2*c)+6*a^2*b^2*sin(d*x+c)+4*a*b^3*(d*x+c)+b^4*ln(sec(d*x+c)+ta n(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.85 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {3 \, b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, {\left (a^{3} b + 2 \, a b^{3}\right )} d x + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{3} b \cos \left (d x + c\right ) + 2 \, a^{4} + 18 \, a^{2} b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]
1/6*(3*b^4*log(sin(d*x + c) + 1) - 3*b^4*log(-sin(d*x + c) + 1) + 12*(a^3* b + 2*a*b^3)*d*x + 2*(a^4*cos(d*x + c)^2 + 6*a^3*b*cos(d*x + c) + 2*a^4 + 18*a^2*b^2)*sin(d*x + c))/d
\[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{4} \cos ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=-\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b - 24 \, {\left (d x + c\right )} a b^{3} - 3 \, b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, a^{2} b^{2} \sin \left (d x + c\right )}{6 \, d} \]
-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3*b - 24*(d*x + c)*a*b^3 - 3*b^4*(log(sin(d*x + c) + 1) - log(s in(d*x + c) - 1)) - 36*a^2*b^2*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {3 \, b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 6 \, {\left (a^{3} b + 2 \, a b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]
1/3*(3*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(a^3*b + 2*a*b^3)*(d*x + c) + 2*(3*a^4*tan(1/2*d*x + 1/ 2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^ 5 + 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a ^4*tan(1/2*d*x + 1/2*c) + 6*a^3*b*tan(1/2*d*x + 1/2*c) + 18*a^2*b^2*tan(1/ 2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 13.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.37 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {3\,a^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {8\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]